Accelerating Around an
Unbanked Curve
Carl E. Mungan, U.S. Naval Academy, Annapolis, MD
T
he December 2004 issue of TPT presented a
problem concerning how a car should accelerate around an unbanked curve of constant
radius r starting from rest if it is to avoid skidding.1
Interestingly enough, two solutions were proffered by
readers.2 The purpose of this note is to compare and
contrast the two approaches. Further experimental
investigation of various turning strategies using a remote-controlled car and overhead video analysis could
make for an interesting student project.
One approach, exemplified by Scott Wiley’s solution, assumes that the tangential acceleration at of the
car is constant throughout the turn, while the centripetal acceleration ac = v 2/r continuously increases as
the car’s speed v builds up. But the magnitude of the
total acceleration a = a t2 + a c2 must never exceed
µs g, where µs is the coefficient of static friction between the tires and road. To minimize the travel time
t, the car’s acceleration just attains this slipping value
as it completes the turn. Using the kinematic relations v 2 = 2 at r φ and r φ = 12 att 2 (where φ is the angle
through which the car has turned), straightforward
algebraic manipulations lead to
v
2φ
=
,
V
k
and
t
= 2 kφ ,
T
ac 2φ
=
,
A
k
at 1
= ,
A k
2
where k ≡ 1 + 4φmax
and the variables have
THE PHYSICS TEACHER ◆ Vol. 44, February 2006
been normalized by V ≡ µs gr , T ≡ r / µs g , and
A µs g. Equations (1) are plotted in red in Fig. 1
for the case of a 45 turn (i.e., φmax = π/4 rad).
The other approach further reduces the driving
time by making a equal to µ s g during the entire turn,
rather than merely as the car completes it. Following the solution of Eugene Mosca, one can equate
at = a 2 − ac2 = V 4 − v 4 r to at = dv/dt = (ds/dt) (dv/ds) = vdv/rdφ (where s = rφ is the distance traveled
around the curve) to obtain
φ=∫
(2)
Inverting this result gives
a
a
v
= sin 2φ ⇒ c = sin 2φ and t = cos 2φ.
A
V
A
(3a)
(If φmax is greater than 45, then v remains constant
with value V, so that ac = A and at = 0, at all angles
beyond π/4.) The time is found by substituting v =
rd φ /dt into the first equality in Eq. (3a) to get
φ
(1)
⎛ v ⎞2
1
= sin−1 ⎜⎜⎜ ⎟⎟⎟ .
⎝V ⎠
V 4 −v4 2
v dv
t
=
T ∫
0
d φ′
sin 2φ ′
≈
φ
N
N
⎡ 2φ ⎛
1 ⎞⎤
∑ csc1/ 2 ⎢⎢ N ⎜⎜⎜⎝n − 2 ⎟⎟⎟⎠⎥⎥,
n=1
⎣
⎦
(3b)
where N is any large integer.3 This summation can
be easily performed in a spreadsheet program.4 The
results are plotted (using N = 1000), along with
Eq. (3a), in blue in Fig. 1.
99
normalized acceleration
normalized speed
1
0.8
0.6
0.4
0.2
(a)
0
1
0.8
0.6
0.4
0.2
(b)
normalized time
0
1.5
1
0.5
(c)
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
angle (rad)
Fig. 1. In each graph, Eq. (1), which assumes constant
tangential acceleration, is plotted in red, and Eq. (3),
which assumes constant total acceleration, is in blue. (a)
The normalized speed v/V of the car. (b) The normalized
tangential acceleration at/A (dashed curves) and centripetal acceleration ac/A (solid curves). (c) The normalized
driving time t/T.
The shapes of the at curves in Fig. 1(b) give an indication of how the driver should depress the gas pedal.
In the first approach (red dashed curve), he needs to
maintain a constant but intermediate pressure on the
pedal. In contrast, in the second scheme (blue dashed
curve) he begins by “flooring it,” such that the tires are
on the verge of slipping, to get the maximum possible
initial increase in speed. As he proceeds around the
100
curve, he then eases off the pedal at an increasing rate
in such a manner that his foot comes fully off it after
having turned through 45 (assuming air drag and
rolling friction are negligible).
As a final remark, one can use the preceding equations to determine how the car travels around the
curve even if it starts with a nonzero speed v0 < V.
(If the car’s initial speed is greater than V, it needs to
first slow down to V and then execute the turn at that
constant speed, as discussed in Ref. 5.) Simply solve
Eq. (1) or (2) as appropriate to find the angle, call
it φ0, that corresponds to the initial speed v0. Now,
rather than performing the subsequent calculations
for angles 0 to φmax, instead consider the automobile
to be traveling from angles φ0 to φ0 + φmax.
References
1. “Physics Challenges for Teachers and Students: A home
stretch,” Phys. Teach. 42, 555 (Dec. 2004).
2. “Solutions to December 2004 Challenges: A home
stretch,” Phys. Teach. 43, 19–20 (2005) online at http://
www.aapt.org/tpt.
3. The factor of ½ in the argument of the cosecant causes
the integrand to be evaluated at the midpoint of each
summation interval of width φ /N . This gives slightly
more accurate results than evaluating the integrand at
the endpoint, because cosecant is a monotonically decreasing function over the range of integration.
Note that the integral in Eq. (3b) for φ = π/4 can be expressed exactly in terms of the gamma function Γ(5 / 4),
whose value can be looked up in standard mathematical
tables. The result is t /T = [ 2Γ(1.25)]2 / 2π ≈ 1.31, in
excellent agreement with the value 1.30 obtained from
the spreadsheet summation. The gamma function is a
generalization of factorials to non-integers. See M.L.
Boas, Mathematical Methods in the Physical Sciences,
2nd ed. (Wiley, New York, 1983), Chap. 11.
5. R.A.D. Hewko, “The racing car turn,” Phys. Teach. 26,
436–437 (Oct. 1988).
PACS codes: 45.40.Aa, 45.20.da, 02.60.-x
4.
Carl Mungan is an assistant professor of physics at the
U.S. Naval Academy with research interests in laser cooling of solids, organic LEDs, and infrared spectroscopy.
Physics Department, U.S. Naval Academy
Annapolis, MD 21402-5002; mungan@usna.edu
THE PHYSICS TEACHER ◆ Vol. 44, February 2006